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Thermal Expansion worksheet

Grades 10–11

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Expansion type: Volumetric (volume)
- Material: Water (near 20 °C): β ≈ 207 × 10⁻⁶ /K
- Starting volume (V₀): 65 L
- Temperature change (ΔT): 50 °C
- Show change in volume in: Cubic meters (m³)
Find: Change in volume (ΔV)
Answer
2.
Problem 2.
- Expansion type: Linear (length)
- Material: Brass: α ≈ 19 × 10⁻⁶ /K
- Starting length (L₀): 24 m
- Temperature change (ΔT): 39 °C
- Show change in length in: Kilometers (km)
Find: Change in length (ΔL)
Answer
3.
Problem 3.
- Expansion type: Linear (length)
- Material: Carbon steel: α ≈ 12 × 10⁻⁶ /K
- Starting length (L₀): 28 m
- Temperature change (ΔT): 40 °C
- Show change in length in: Inches (in)
Find: Change in length (ΔL)
Answer
4.
Problem 4.
- Expansion type: Linear (length)
- Material: Invar: α ≈ 1.2 × 10⁻⁶ /K
- Starting length (L₀): 26 m
- Temperature change (ΔT): 30 °C
- Show change in length in: Miles (mi)
Find: Change in length (ΔL)
Answer
5.
Problem 5.
- Expansion type: Volumetric (volume)
- Material: Ethanol: β ≈ 1,100 × 10⁻⁶ /K
- Starting volume (V₀): 50 L
- Temperature change (ΔT): 50 °C
- Show change in volume in: Cubic centimeters (cm³)
Find: Change in volume (ΔV)
Answer
6.
Problem 6.
- Expansion type: Volumetric (volume)
- Material: Enter a coefficient
- Volumetric expansion coefficient (β): 275 ×10⁻⁶ /K
- Starting volume (V₀): 42 L
- Temperature change (ΔT): 55 °C
- Show change in volume in: US gallons (US gal)
Find: Change in volume (ΔV)
Answer
7.
Problem 7.
- Expansion type: Linear (length)
- Material: Borosilicate glass: α ≈ 3.3 × 10⁻⁶ /K
- Starting length (L₀): 28 m
- Temperature change (ΔT): 32 °C
- Show change in length in: Miles (mi)
Find: Change in length (ΔL)
Answer
8.
Problem 8.
- Expansion type: Volumetric (volume)
- Material: Glycerin: β ≈ 500 × 10⁻⁶ /K
- Starting volume (V₀): 58 L
- Temperature change (ΔT): 40 °C
- Show change in volume in: Milliliters (mL)
Find: Change in volume (ΔV)
Answer
9.
Problem 9.
- Expansion type: Volumetric (volume)
- Material: Glycerin: β ≈ 500 × 10⁻⁶ /K
- Starting volume (V₀): 30 L
- Temperature change (ΔT): 40 °C
- Show change in volume in: Liters (L)
Find: Change in volume (ΔV)
Answer
10.
Problem 10.
- Expansion type: Volumetric (volume)
- Material: Ethanol: β ≈ 1,100 × 10⁻⁶ /K
- Starting volume (V₀): 65 L
- Temperature change (ΔT): 22 °C
- Show change in volume in: Cubic feet (ft³)
Find: Change in volume (ΔV)
Answer

Answer key: Thermal Expansion worksheet

1. Change in volume (ΔV): 6.73e-4 m³
1. Look up the typical expansion rate for water (near 20 °C): how much of its size it grows per degree
  $β$ $= 207 \times 10^{−6}$ $= 0.000207 /K$
2. Multiply the expansion rate by the starting volume and the temperature change to get the growth
  $0.000207 \times 65 \times 50$ $= 0.67275 L$
3. Change the result into the unit you picked, m³ (1 L = 0.001 m³)
  $0.67275 L \times 0.001$ $= 0.000673 m³$
4. Add the change to the starting volume to get the new volume (V = V₀ + ΔV)
  $65 + 0.67275$ $= 65.67275 L$
2. Change in length (ΔL): 1.78e-5 km
1. Look up the typical expansion rate for brass: how much of its size it grows per degree
  $α$ $= 19 \times 10^{−6}$ $= 0.000019 /K$
2. Multiply the expansion rate by the starting length and the temperature change to get the growth
  $0.000019 \times 24 \times 39$ $= 0.017784 m$
3. Change the result into the unit you picked, km (1 m = 0.001 km)
  $0.017784 m \times 0.001$ $= 0.000018 km$
4. Add the change to the starting length to get the new length (L = L₀ + ΔL)
  $24 + 0.017784$ $= 24.017784 m$
3. Change in length (ΔL): 0.53 in
1. Look up the typical expansion rate for carbon steel: how much of its size it grows per degree
  $α$ $= 12 \times 10^{−6}$ $= 0.000012 /K$
2. Multiply the expansion rate by the starting length and the temperature change to get the growth
  $0.000012 \times 28 \times 40$ $= 0.01344 m$
3. Change the result into the unit you picked, in (1 m = 39.370078740157 in)
  $0.01344 m \times 39.370078740157$ $= 0.529134 in$
4. Add the change to the starting length to get the new length (L = L₀ + ΔL)
  $28 + 0.01344$ $= 28.01344 m$
4. Change in length (ΔL): 5.82e-7 mi
1. Look up the typical expansion rate for invar: how much of its size it grows per degree
  $α$ $= 1.2 \times 10^{−6}$ $= 0.0000012 /K$
2. Multiply the expansion rate by the starting length and the temperature change to get the growth
  $0.0000012 \times 26 \times 30$ $= 0.000936 m$
3. Change the result into the unit you picked, mi (1 m = 0.000621371192 mi)
  $0.000936 m \times 0.000621371192$ $= 5.816034e-7 mi$
4. Add the change to the starting length to get the new length (L = L₀ + ΔL)
  $26 + 0.000936$ $= 26.000936 m$
5. Change in volume (ΔV): 2,750 cm³
1. Look up the typical expansion rate for ethanol: how much of its size it grows per degree
  $β$ $= 1,100 \times 10^{−6}$ $= 0.0011 /K$
2. Multiply the expansion rate by the starting volume and the temperature change to get the growth
  $0.0011 \times 50 \times 50$ $= 2.75 L$
3. Change the result into the unit you picked, cm³ (1 L = 1,000 cm³)
  $2.75 L \times 1,000$ $= 2,750 cm³$
4. Add the change to the starting volume to get the new volume (V = V₀ + ΔV)
  $50 + 2.75$ $= 52.75 L$
6. Change in volume (ΔV): 0.17 US gal
1. Change your expansion rate to per kelvin, the unit the formula uses
  $275 \times 0.000001$ $= 0.000275 /K$
2. Multiply the expansion rate by the starting volume and the temperature change to get the growth
  $0.000275 \times 42 \times 55$ $= 0.63525 L$
3. Change the result into the unit you picked, US gal (1 L = 0.264172052358 US gal)
  $0.63525 L \times 0.264172052358$ $= 0.167815 US gal$
4. Add the change to the starting volume to get the new volume (V = V₀ + ΔV)
  $42 + 0.63525$ $= 42.63525 L$
7. Change in length (ΔL): 1.84e-6 mi
1. Look up the typical expansion rate for borosilicate glass: how much of its size it grows per degree
  $α$ $= 3.3 \times 10^{−6}$ $= 0.0000033 /K$
2. Multiply the expansion rate by the starting length and the temperature change to get the growth
  $0.0000033 \times 28 \times 32$ $= 0.0029568 m$
3. Change the result into the unit you picked, mi (1 m = 0.000621371192 mi)
  $0.0029568 m \times 0.000621371192$ $= 0.000002 mi$
4. Add the change to the starting length to get the new length (L = L₀ + ΔL)
  $28 + 0.0029568$ $= 28.0029568 m$
8. Change in volume (ΔV): 1,160 mL
1. Look up the typical expansion rate for glycerin: how much of its size it grows per degree
  $β$ $= 500 \times 10^{−6}$ $= 0.0005 /K$
2. Multiply the expansion rate by the starting volume and the temperature change to get the growth
  $0.0005 \times 58 \times 40$ $= 1.16 L$
3. Change the result into the unit you picked, mL (1 L = 1,000 mL)
  $1.16 L \times 1,000$ $= 1,160 mL$
4. Add the change to the starting volume to get the new volume (V = V₀ + ΔV)
  $58 + 1.16$ $= 59.16 L$
9. Change in volume (ΔV): 0.6 L
1. Look up the typical expansion rate for glycerin: how much of its size it grows per degree
  $β$ $= 500 \times 10^{−6}$ $= 0.0005 /K$
2. Multiply the expansion rate by the starting volume and the temperature change to get the growth
  $0.0005 \times 30 \times 40$ $= 0.6 L$
3. Add the change to the starting volume to get the new volume (V = V₀ + ΔV)
  $30 + 0.6$ $= 30.6 L$
10. Change in volume (ΔV): 0.06 ft³
1. Look up the typical expansion rate for ethanol: how much of its size it grows per degree
  $β$ $= 1,100 \times 10^{−6}$ $= 0.0011 /K$
2. Multiply the expansion rate by the starting volume and the temperature change to get the growth
  $0.0011 \times 65 \times 22$ $= 1.573 L$
3. Change the result into the unit you picked, ft³ (1 L = 0.035314666721 ft³)
  $1.573 L \times 0.035314666721$ $= 0.05555 ft³$
4. Add the change to the starting volume to get the new volume (V = V₀ + ΔV)
  $65 + 1.573$ $= 66.573 L$